1
$\begingroup$

I was thinking of an example of a matrix $A$ which is at least $4$ dimensional, that has no real eigenvalue but yet det$(\Phi) = 0$, where the matrix $\Phi$ is given by $\begin{bmatrix} e_{1}^{T} & \\ e_{1}^{T}A \\ e_{1}^{T}A^2 \\ .\\ .\\ .\\ e_{1}^{T}A^{n-1}\end{bmatrix}$

Where $e_{1}$ is the standard basis vector of first entry 1 and others 0.

The eigenvalues of $A$ are complex so they must occur in pairs and hence I think we, therefore, need for $A$ to be even-dimensional and must be greater than or equal to $4$, also the determinant is the product of the eigenvalues will be a positive number then as we will be multiplying the complex eigenvalue and its conjugate, so the determinant will be positive but I am thinking now how the determinant can be zero? as there is no eigenvalue 0.

$\endgroup$
  • 1
    $\begingroup$ If a matrix has zero determinant, then $0$ is an eigenvalue, and $0$ is a real number. $\endgroup$ – Rellek Jun 21 '18 at 12:31
  • $\begingroup$ Yup, but $A$ has no real eigenvalues!, but yes it would be sufficient to show that matrix $\Phi$ has a zero eigenvalue!! so now we have to think how to prove that the matrix $\Phi$ has a zero eigenvalue? $\endgroup$ – BAYMAX Jun 21 '18 at 12:35
  • $\begingroup$ That's what I said. If a matrix is singular, then the nullspace is nontrivial. Thus any generator of the nullspace is an eigenvector for the eigenvalue $0$. $\Phi$ is singular, so $0$ is an eigenvalue. $\endgroup$ – Rellek Jun 21 '18 at 12:38
  • $\begingroup$ What is $e_1$? Is it really the same for every row? $\endgroup$ – Aleksejs Fomins Jun 21 '18 at 16:25
  • $\begingroup$ Your matrix has $n+1$ rows, but (presumably) $n$ columns. Then it has no determinant. $\endgroup$ – Marc van Leeuwen Jun 21 '18 at 16:34
0
$\begingroup$

The determinant asked about is not that of $A$, but that of repeated (right) images of the first row $e_1^T$ of the identity matrix. If you take $A$ to be a block diagonal matrix, then those images remain "inside the first block", so they will never produce a full rank matrix, and the determinant will be$~0$. So take a block diagonal matrix of at least two $2\times 2$ rotation matrices, which have no real eigenvalues, like $$ A=\pmatrix{0&1&0&0\\-1&0&0&0\\0&0&0&-1\\0&0&1&1} $$ which gives $$ \Phi=\pmatrix{1&0&0&0\\0&1&0&0\\-1&0&0&0\\0&-1&0&0}, $$ obviously singular.

$\endgroup$
  • $\begingroup$ How repeated images of 1st row of identity matrix i think its like first row is the transpose of e1 vector and 2nd row is the first row of A nd 3rd row is 1st row of A^2 nd so on... I could not understand the images you mention after that, it would be much helpful if you could elaborate this and with a motivation of why we are taking A to be block diagknal matrix.. $\endgroup$ – BAYMAX Jun 22 '18 at 12:55
  • $\begingroup$ @BAYMAX These are "repeated images", since $e_1^TA^{k+1}=(e_1^TA^k)A$. The motivation for using a bock diagonal matrix is that you want to lock up the repeated images in an invariant subspace (of the row space) to avoid them getting to span the whole space, and at the same time it allows easy control of the eigenvalues. $\endgroup$ – Marc van Leeuwen Jun 26 '18 at 4:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.